Chapter 4
The Mole and
Stoichiometry
Chemistry: The Molecular Nature
of Matter, 6E
Brady/Jespersen/Hyslop
• The molecular scale versus the
laboratory scale:
• Defining the mole:
• A number equal to the number of atoms
in exactly 12 gram of 𝟏𝟐𝑪 atoms.
• 1 mole of element X= gram atomic mass of X
• Example:
• 1 mole of sulfur = 32.6 g
atomic mass of sulfur = 32.6 u
• The mole concept applied to compound:
• 1 mole of molecules X = gram molecular
mass of X
• Example:
• The molecular mass of water = 18.02 u
• The sum of atomic mass of two H atoms
and one O atmo.
• 1 mole of 𝑯𝟐 O= 18.02 g
• 1 mole of ionic compound X = gram
formula mass of X
• Example:
• 1 mole of 𝑨𝑳𝟐 𝑶𝟑 =101.96 g
• 𝑨𝑳𝟐 𝑶𝟑 has 2AL with an atomic masses
26.98u and three oxygen with mass of
16 u.
• 1 mole of X = gram molar mass of X
• Converting from gram to moles:
• Example:
• In experiment to prepare of titanium(IV)
oxide we start with 23.5g sample of
titanium. How many moles of titanium do
we have?
• Solution:
• 1 mole of element X = gram atomic mass of
X
• 1 mole of Ti = 47.867g Ti
• 23 .5g Ti = ? Mole Ti
• Conversion factor:
𝟏 𝒎𝒐𝒍𝒆 𝑻𝒊
𝟒𝟕. 𝟖𝟔𝟕 𝒈𝑻𝒊
• 23.5 gTi ×
𝟏 𝒎𝒐𝒍𝒆 𝑻𝒊
𝟒𝟕,𝟖𝟔𝟕 𝒈𝑻𝒊
= 0.491 mole Ti
• Conversion from mole to grams:
• Example:
• We need 0.254 moles of 𝑭𝒆𝑪𝑳𝟑 for
certain experiment. How many grams
would you need weight?
• Solution:
• 1 mole of 𝑭𝒆𝑪𝑳𝟑 = 162.204 g
• Molar mass 𝑭𝒆𝑪𝑳𝟑 = 55.845 g/mole +
( 3× 35.453) g/mole = 162.204 g/mole
• 0.254 mole 𝑭𝒆𝑪𝑳𝟑 = ? g 𝑭𝒆𝑪𝑳𝟑
• Conversion factor:
𝟏𝟔𝟐.𝟐𝟎𝟒 𝒈𝑭𝒆𝑪𝑳𝟑
𝟏 𝒎𝒐𝒍𝒆 𝑭𝒆𝑪𝑳𝟑
0.254 ×
𝟏𝟔𝟐.𝟐𝟎𝟒 𝒈𝑭𝒆𝑪𝑳𝟑
𝟏 𝒎𝒐𝒍𝒆 𝑭𝒆𝑪𝑳𝟑
= 41.2 g 𝑭𝒆𝑪𝑳𝟑
Learning check
* How many moles of aluminum are
there iN 3.47 gram sheet of aluminum foil.
- Atomic mass of AL=26.98u
• Avogadros number:
• The relationship between the atomic scale
and laboratory scale as
• 1 mole of X = 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 unit of X
• The unit can be atoms , molecules , formula
unit.
• Example :
• 1 mole of Xe = 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 atoms of Xe
• 1 mole of 𝑵𝑶𝟐 = 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 molecules
of 𝑵𝑶𝟐
• Converting from the laboratory scale to
the atomic scale.
• Example:
• In lightbulb the tungsten weight 0.653 g.
how many atoms of tungsten are there
in such sample.
• Solution:
• 0.632 g W = ? Atoms of W
• gram W to mole W to atom W
• 1 mole of W = 183.84 g W
• 1 mole W = 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 atoms W
• Conversion factor :
• 1.
𝟏 𝒎𝒐𝒍𝒆 𝑾
𝟏𝟖𝟑.𝟖𝟒 𝒈 𝑾
• 2.
𝟔.𝟎𝟐𝟐×𝟏𝟎𝟐𝟑 𝒂𝒕𝒐𝒎𝒔 𝑾
𝟏 𝒎𝒐𝒍𝒆 𝑾
• 0.635 ×
𝟏 𝒎𝒐𝒍𝒆 𝑾
𝟏𝟖𝟑.𝟖𝟒 𝒈 𝑾
×
𝟔.𝟎𝟐𝟐×𝟏𝟎𝟐𝟑 𝒂𝒕𝒐𝒎𝒔 𝑾
𝟏 𝒎𝒐𝒍𝒆 𝑾
𝟐𝟏
2.08 × 𝟏𝟎
=
atoms W
• Calculating the mass of molecules:
• Example :
• What is the average mass of one molecule of
carbon tetrachloride.
• Soluation:
• 1 molecule 𝑪𝑪𝑳𝟒 = ? g 𝑪𝑪𝑳𝟒
• 1 mole of 𝑪𝑪𝑳𝟒 = 𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 molecules of
𝑪𝑪𝑳𝟒
• 1 mole of 𝑪𝑪𝑳𝟒 = 153.823 g 𝑪𝑪𝑳𝟒
• Molar mass of 𝑪𝑪𝑳𝟒 = 12 + (4×
35.45)=153.823 g\mole
• Conversion factor :
• 1.
𝟏 𝒎𝒐𝒍𝒆 𝑪𝑪𝑳𝟒
𝟔.𝟎𝟐𝟐×𝟏𝟎𝟐𝟑 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆 𝑪𝑪𝑳𝟒
• 2.
𝟏𝟓𝟑.𝟖𝟐𝟑 𝒈𝑪𝑪𝑳𝟒
𝟏 𝒎𝒐𝒍𝒆 𝑪𝑪𝑳𝟒
• 1 molecule 𝑪𝑪𝑳𝟒 ×
𝟏 𝒎𝒐𝒍𝒆 𝑪𝑪𝑳𝟒
𝟏𝟓𝟑.𝟖𝟐𝟑 𝒈𝑪𝑪𝑳𝟒
×
𝟐𝟑
𝟔.𝟎𝟐𝟐×𝟏𝟎 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆 𝑪𝑪𝑳𝟒
𝟏 𝒎𝒐𝒍𝒆 𝑪𝑪𝑳𝟒
𝟐. 𝟓𝟓𝟓 × 𝟏𝟎−𝟐𝟐 g 𝑪𝑪𝑳𝟒 .
=
Learning check
• A sample of ethanol contains 1.6 mol C2H5OH,
Calculate the mass of C2H5OH molecules in
this sample. ( molar mass is , C = 12 , H=1 ,
O = 16 g/mol ).
• Calculate the mole number of 15.055 X 1023
CO molecules. ( Avogadro's number = 6.022 x
1023).
• Percentage composition:• Called percentage by mass of element:
is the number of grams of the element
present in 100g of the compound.
• Percentage by mass of element =
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒆𝒍𝒆𝒎𝒆𝒏𝒕
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒘𝒉𝒐𝒍𝒆 𝒔𝒂𝒎𝒑𝒍𝒆
× 𝟏𝟎𝟎%
• Example:
• A sample of liquid with a mass of 8.657 g
was decomposed into its element and gave
5.217g of carbon, 0.9620 g of hydrogen, and
2.478 g of oxygen. What is the percentage
composition of this compound?
• solution:
• For C :
• For H :
𝟓.𝟐𝟏𝟕 𝒈
× 𝟏𝟎𝟎% = 𝟔𝟎. 𝟐𝟔%
𝟖.𝟔𝟓𝟕 𝒈
𝟎.𝟗𝟔𝟐 𝒈
× 𝟏𝟎𝟎% = 𝟏𝟏. 𝟏𝟏 %
𝟖.𝟔𝟓𝟕 𝒈
C
• For O :
𝟐.𝟒𝟕𝟖 𝒈
𝟖.𝟔𝟓𝟕 𝒈
× 𝟏𝟎𝟎 % = 𝟐𝟖. 𝟔𝟐 %
• Sum of percentage: 99.99%
Learning check
* An organic compound weighing 0.6672 g
is decomposed , giving 0.3481 g of carbon
0,087 g hydrogen. What is the percentage
of hydrogen and carbon in this
compound?
Lreaning check
• Calculate the percentage composition of each
element in KNO3 (Atomic masses ,K = 39 ,
N = 14 , O = 16 g/mol).
• Determining empirical and molecular
formulas:
• Calculating an empirical formula from
mass data:
• Example:
• 2.57 g sample of compound composed on only tin
and chlorine was found to contain 1.17 g of tin.
What the compound empirical formula?
•
• Solution:
• Mass of CL = 2.57 g compound – 1.17 g Sn =
1.40 g CL
• 1 mol Sn = 118.7 g Sn
• 1 mol CL = 35.45 g CL
• C.V :
• 1.
1 𝑚𝑜𝑙 𝑆𝑛
118.7 𝑔 𝑆𝑛
• 2.
1 𝑚𝑜𝑙 𝐶𝐿
35.45 𝑔 𝐶𝐿
• 1.17g ×
1 𝑚𝑜𝑙 𝑆𝑛
118.7 𝑔 𝑆𝑛
• 1.40 g ×
= 0.00986 mol Sn
1 𝑚𝑜𝑙 𝐶𝐿
35.45 𝑔 𝐶𝐿
= 0.0395 mol CL
• Formula: 𝑆𝑛0.00986 𝐶𝐿0.0395
• To convert the decimal subscripts to integer by
dividing each by the smallest number in the
set .
• 𝑆𝑛 0.00986 𝐶𝐿 0.0395 = 𝑆𝑛1.00 𝐶𝐿4.01
0.00986
0.00986
• The empirical formula is Sn𝐶𝑙4
Learning check
A 1.525g sample of a compound between
nitrogen and oxygen contain 0.712g of
nitrogen. Calculate its empirical formula.
Determining a molecular formula from an
empirical formula and a molecular mass:𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅
𝒆𝒎𝒑𝒊𝒓𝒊𝒄𝒂𝒍 𝒇𝒐𝒓𝒎𝒖𝒍𝒂 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒄𝒐𝒎𝒑𝒐𝒖𝒏𝒅
= integer
Example:Styrene has an empirical formula of CH it molecular
mass is 104. what is its molecular formula?
Solution:
The formula mass is
12.01 + 1.008 = 13.02
𝟏𝟎𝟒
𝟏𝟑.𝟎𝟐
= 7.99 = 8
Molecular formula of styrene is 𝑪𝟖 𝑯𝟖
learning check
The empirical formula of hydrazine is
N𝑯𝟐 and its molecular mass is 32.0
what is its molecular formula?
• The mole and chemical reaction:• Writing and balancing equations:
• Always the balancing of an equation as a
two-step process:
• Step 1: write the unbalanced equation.
• Step 2: adjust the coefficient to get equal
number of each kind of atoms on both side.
• Some guideline for balancing equation:
• 1. start balancing with the most complicated
formula first . Element, particularly
𝑯𝟐 𝒂𝒏𝒅 𝑶𝟐 .
• 2. balance atoms that appear in only two
formula.
• 3. balance as a group those polyatomic ions
that appear un changed on both side of the
arrow.
• Example:• Sodium hydroxide and phosphoric acid
𝑯𝟑 𝑷𝑶𝟒 React to give sodium phosphate and
water. The sodium posphate remain in
solution. Write the balanced equation for this
reaction?
• Solution:
• NaOH + 𝑯𝟑 𝑷𝑶𝟒
𝑵𝒂𝟑 𝑷𝑶𝟒 + 𝑯𝟐 O
(unblanced)
•
•
•
•
1. balance element ( Na and P)
3 NaOH + 𝑯𝟑 𝑷𝑶𝟒
𝑵𝒂𝟑 𝑷𝑶𝟒 + 𝑯𝟐 O
2. balance particular ( 𝑯𝟐 − 𝑶𝟐 )
3 NaOH + 𝑯𝟑 𝑷𝑶𝟒
𝑵𝒂𝟑 𝑷𝑶𝟒 + 𝟑 𝑯𝟐 O
LEARNING CHECK
Balance the following equation:
1. Ca𝑪𝑳𝟐 + 𝑲𝟑 𝑷𝑶𝟒
𝑪𝒂𝟑 (𝑷𝑶𝟒 )2 + KCL
2. 𝑲𝟐 𝑺𝑶𝟒 + NaOH
𝑵𝒂𝟐 𝑺𝑶𝟒 + KOH
• Example :
• How many moles of sodium phosphate can be
made from 0.24 mol of sodium hydroxide by the
following equation:
• 3NaOH (aq) + H3PO4(aq) → Na3PO4(aq) + 3 H2O
sol:
3 mol of NaOH ↔ 1mol of Na3PO4
0.24 mol NaOH ↔ ? mol of Na3PO4
C.V :
1mol of Na3PO4
3 mol of NaO𝐻
• 0.24 mol NaOH × 1mol of Na3PO4 =
3 mol of NaO𝐻
0.08 mol Na3PO4
Learning chick
• How many moles of O2 are needed to
produced 6.76 moles of SO3 from the
following equation:
2SO2 + O2 → 2SO3